0 push 1 1 duplicate top of stack 2 add top two values 3 subtract 4 mutliply 5 integer divide top value byb second value 6 push the second value without popping it 7 swap top two values c>7 while loop, runs until top of stack is 0 the loops ends at the first instruction >=c 

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0 push 1 1 duplicate top of stack 2 add top two values 3 subtract 4 multiply 5 integer divide top value by second value 6 push the second value without popping it 7 swap top two values c>7 while loop, runs until top of stack is 0 the loops ends at the first instruction >=c 

0,20,10,0,3,10,2,0,3,comp. test,20,2,1 generates the next prime less than n:

0 -- push 1. This means the current number is composite -- Even if it isn't, we still want to find a prime <n 20 -- while loop. This iterates until the composite tests returns 0 10 10 -- we have an positive number on the top of the stack ... 0 3 -- by subtracting 1 until it is 0, ... 2 -- and adding this to the last prime candidate ... -- we can get rid of it. 0 3 -- subtract 1 to get new prime candidate pc comp -- check if pc is composite 20 -- end of loop, top of stack is now [0, p], with p prime 2 -- add 0+p 1 -- duplicate the prime, such that we store the result, -- and can use the value to find the next prime 

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30,0,2,0,2,next prime,0,3,0,3,30,2 repeats this until the prime 2 is found:

30 30 -- while loop 0 2 -- add 1 0 2 -- add 1 np -- find the prime less than this 0 3 -- subtract 1 0 3 -- subtract 1 -- if the prime was 2, this is now 0 -- and the while loop terminates 2 -- add the 0 to the 2 to remove it 

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The available operations were

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The available operations are

0,0,2,1,4,1,0,2,1,1,4,4,4 pushes the initial number 500. The remainder of the code is best explained inside out:

6,6,6,6,5,4,7,5 is a divisibility test. Given k and n as the top two values on the stack this calculates \$\lfloor {\lfloor {n \over k} \rfloor \cdot k \over n}\rfloor\$, which is only 1 if k divides n: Try it online!

1,0,10,3,div test,0,7,3,10,2,0,3 is a primality test, or a composite test since this returns truthy (non-zero) values for composite numbers:

1 -- duplicate n 0 -- push 1 - stack: [1, k=n, n] -- in the next iterations of the loop, -- the top of stack will be the inverted result -- of the divisibility test 10 10 -- while loop: -- runs until [0, d, n] is on the stack, -- where d is the largest divisor of n <n 3 -- subtract top value (always 1) from k div -- the divisibility test 0 -- push 1 7 -- swap top two values 3 -- subtract (1 - div test result) 2 -- add the top 0 to the last k 0,3 -- subtract 1 -- if the loop ended with [0, 1, n], this is now 0 -- otherwise we have a positive number 

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